hdu 4035 Maze 在树上求逃出去的期望

Problem Description

When wake up, lxhgww find himself in a huge maze.

The maze consisted by N rooms and tunnels connecting these rooms. Each pair of rooms is connected by one and only one path. Initially, lxhgww is in room 1. Each room has a dangerous trap. When lxhgww step into a room, he has a possibility to be killed and restart from room 1. Every room also has a hidden exit. Each time lxhgww comes to a room, he has chance to find the exit and escape from this maze.

Unfortunately, lxhgww has no idea about the structure of the whole maze. Therefore, he just chooses a tunnel randomly each time. When he is in a room, he has the same possibility to choose any tunnel connecting that room (including the tunnel he used to come to that room).
What is the expect number of tunnels he go through before he find the exit?

 

Input

First line is an integer T (T ≤ 30), the number of test cases.

At the beginning of each case is an integer N (2 ≤ N ≤ 10000), indicates the number of rooms in this case.

Then N-1 pairs of integers X, Y (1 ≤ X, Y ≤ N, X ≠ Y) are given, indicate there is a tunnel between room X and room Y.

Finally, N pairs of integers Ki and Ei (0 ≤ Ki, Ei ≤ 100, Ki + Ei ≤ 100, K1 = E1 = 0) are given, indicate the percent of the possibility of been killed and exit in the ith room.

 

Output

For each test case, output one line “Case k: ”. k is the case id, then the expect number of tunnels lxhgww go through before he exit. The answer with relative error less than 0.0001 will get accepted. If it is not possible to escape from the maze, output “impossible”.

 

Sample Input

   
   
   
   

    
    
    
    3
3
1 2
1 3
0 0
100 0
0 100
3
1 2
2 3
0 0
100 0
0 100
6
1 2
2 3
1 4
4 5
4 6
0 0
20 30
40 30
50 50
70 10
20 60
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1: 2.000000
Case 2: impossible
Case 3: 2.895522
   
   
   
   

//

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=22000;
//dp[i]=ki*dp[1]+pet*(dp[j]+1)+pet*(dp[fath]+1)
//从底向上 层层递推
//最上层解出dp[1]即可
struct Node
{
    int t;
    int flag;
    int next;
};
int n;
double k[maxn],e[maxn];
double c[maxn],x[maxn],f[maxn];//常数系数和dp1系数 fath的系数
int p[maxn];
int l;
Node G[maxn];
void init()
{
    memset(p,-1,sizeof(p));
    l=0;
}
void addedge(int u,int t)
{
    G[l].t=t;
    G[l].flag=0;
    G[l].next=p[u];
    p[u]=l++;
}


void dfs(int u,int fath)
{
    c[u]=0;x[u]=k[u]/100;
    int cnt=0;
    for(int i=p[u];i!=-1;i=G[i].next)
    {
        int t=G[i].t;
        cnt++;
    }
    double pet;
    pet=(100-k[u]-e[u])/cnt/100;
    if(u!=1) f[u]=pet,c[u]+=pet;
    double res=0;
    for(int i=p[u];i!=-1;i=G[i].next)
    {
        int t=G[i].t;
        if(t==fath) continue;
        dfs(t,u);
        c[u]+=pet*(1+c[t]);x[u]+=pet*x[t];
        res+=pet*f[t];
    }
    c[u]/=(1-res),x[u]/=(1-res),f[u]/=(1-res);
}
int main()
{
    int ci,pl=1;scanf("%d",&ci);
    while(ci--)
    {
        scanf("%d",&n);
        init();
        for(int i=0;i<n-1;i++)
        {
            int u,t;scanf("%d%d",&u,&t);
            addedge(u,t);
            addedge(t,u);
        }
        for(int i=1;i<=n;i++) scanf("%lf%lf",&k[i],&e[i]);
        //1树根
        memset(c,0,sizeof(c));
        memset(x,0,sizeof(x));
        memset(f,0,sizeof(f));
        dfs(1,-1);
        printf("Case %d: ",pl++);
        if(fabs(x[1]-1)<1e-10) printf("impossible\n");//-8不够 得-10
        else printf("%.6lf\n",c[1]/(1-x[1]));
    }
    return 0;
}